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Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It

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Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It [#permalink]

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New post 09 Feb 2011, 09:14
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Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It continues in such a way that by adding 5 to the nth term, one obtains the (n + 2)th term. What is the sum of the first 19 members of this sequence?

A. 878
B. 900
C. 788
D. 928
E. 1022
[Reveal] Spoiler: OA

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Re: Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It [#permalink]

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New post 09 Feb 2011, 10:12
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I would split them up like this.
23 28 33 and so on (10 terms)....the 10th term = 23+9*5 = 68
27 32 37 and so on (9 terms)......the 9th term = 27+8*5 = 67

Since the distance between any two numbers is the same we can use arithmetics

first+last/2 times no of numbers = the sum

(23+68)/2 * 10 = 455
(27+67)/2 * 9 = 423

= 878
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Re: Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It [#permalink]

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New post 09 Feb 2011, 16:32
Thanks Mackieman! I did in the same way but got screwed as i included everything into the same formula n/2 (2a+ (n-1) d )... it's the same approach
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Re: Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It [#permalink]

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New post 09 Feb 2011, 16:37
bhandariavi wrote:
Thanks Mackieman! I did in the same way but got screwed as i included everything into the same formula n/2 (2a+ (n-1) d )... it's the same approach


Ah okay :). Propably some of the Math Gurus know if there is a simpler and faster approach.

"My" method shouldn't take more than 1-2 minutes though.
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Re: Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It [#permalink]

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Re: Consider the sequence of numbers beginning 23, 27, 28, 32, 33... It   [#permalink] 04 Oct 2017, 03:10
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